Sometimes the limit does not exist and sometimes the it diverges to infinity
Here is an example: $$\lim_{t\to\infty}\ln(\cos t)dt $$ Why does it not exist? I think its infinity, isn't it?
And here $$\lim_{t\to\pi/2}\ln (\sec t+\tan t) =\infty$$ Thanks all
The limit of $f(t)$ as $t \to t_0$ "goes to infinity" if, for all $x \in \mathbb{R}$, getting sufficiently close to $t_0$ makes your outputs larger than $x$.
More formally, if $t_0$ is finite: $$\lim_{t \to t_0} f(t) = \infty \iff \forall x \in \mathbb{R}, \ \exists \delta > 0 \ \left( 0 < |t - t_0| < \delta \implies f(t) > x \right)$$
If $t \to \infty$: $$\lim_{t \to \infty} f(t) = \infty \iff \forall x \in \mathbb{R}, \exists y \in \mathbb{R} \ \left( t > y \implies f(t) > x \right)$$
For your first example, notice that as $t \to \infty$, $\cos t$ can reach anywhere between $-1$ and $1$. No matter what $y$ you pick, there will be a $t_1 > y$ such that $\cos t_1 = 1$ and $t_2 > y$ such that $\cos t_2 = 1/e$. So your function will hit $\ln 1 = 0$ and $\ln 1/e = -1$ no matter how far out you look. It neither "settles down" to any finite number, nor does it "zoom off to infinity".
As for the second, as $t \to \pi/2$, $\sec t$ gets arbitrarily large, but for $\tan t$, it depends on which side it is approached from. So let's rewrite $\sec t + \tan t$ as $\frac{\sin t + 1}{\cos t}$. The numerator is always positive but the denominator changes sign at $\pi/2$. On the left it goes to $\infty$, but on the right, $-\infty$. So this limit does not exist!
However, I suspect there are supposed to be absolute value bars around it, so let's consider that case. Now, we know that $\left| \frac{\sin t + 1}{\cos t} \right|$ goes to $\infty$, that is, for all $x$, we can find an interval around $\pi/2$ such that $f(t) > x$. To show that $\ln |\sec t + \tan t|$ goes to infinity, note that the natural log is monotonically increasing. So to get above a bound of $x$, find an interval around $\pi/2$ such that $|\sec t + \tan t| > e^x$. This implies that $\ln | \sec t + \tan t | > \ln e^x = x$, so we can make that expression arbitrarily large.